Answer
The second ionization energy of $Li$ is much larger than that of $Be$ because the second ionization of $Li$ would remove an inner electron in the $n=1$ shell, while the second ionization of $Be$ would still remove an outer electron in the $n=2$ shell.
Work Step by Step
$Li$ has 1 electron in the outermost shell $(n=2)$, while $Be$ has 2 electrons in the outermost shell $(n=2)$.
The second ionization removes another electron from an ion that has already lost 1 electron.
So, the second ionization of $Li$ would remove an inner electron in the $n=1$ shell, while that of $Be$ would still remove an outer electron in the $n=2$ shell.
Since $n=1$ shell is much closer to the nucleus than $n=2$ shell, an electron in the $n=1$ shell has a higher chance of being closer to the nucleus than that in the $n=2$ shell. An electron that is closer to the nucleus undergoes a much greater attraction from the nucleus than one further from the nucleus.
That means to remove a $n=1$ shell electron requires a much higher energy than to remove a $n=2$ shell electron. That is why the second ionization energy of $Li$ is much larger than that of $Be$.